Problem: $h(t) = 6t^{2}+3t+3(g(t))$ $g(x) = -2x^{2}-x$ $ g(h(8)) = {?} $
Explanation: First, let's solve for the value of the inner function, $h(8)$ . Then we'll know what to plug into the outer function. $h(8) = 6(8^{2})+(3)(8)+3(g(8))$ To solve for the value of $h$ , we need to solve for the value of $g(8)$ $g(8) = -2(8^{2})-8$ $g(8) = -136$ That means $h(8) = 6(8^{2})+(3)(8)+(3)(-136)$ $h(8) = 0$ Now we know that $h(8) = 0$ . Let's solve for $g(h(8))$ , which is $g(0)$ $g(0) = -2(0^{2})-0$ $g(0) = 0$